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sritama
 3 years ago
a1,a2,a3....a30each of these 30 sets has 5 elements.b1,b2,....bneach of these n sets has 3 elements.union of a1,a2...a30=union of b1,b2....bn=S.
if each elements of S is in 10 a sets and 9 bsets.then n=?
ans=45.
show the process
sritama
 3 years ago
a1,a2,a3....a30each of these 30 sets has 5 elements.b1,b2,....bneach of these n sets has 3 elements.union of a1,a2...a30=union of b1,b2....bn=S. if each elements of S is in 10 a sets and 9 bsets.then n=? ans=45. show the process

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joemath314159
 3 years ago
Best ResponseYou've already chosen the best response.1ok, here we go~ So the number of elements in the union of the A sets is: \[5(30)r_A\]where r is the number of repeats. Likewise the number of elements in the B sets is: \[3nr_B\] Each element in the union (in S) is repeated 10 times in A, which means if x was the real number of elements in A (not counting repeats) then 9 out of those 10 should be thrown away, or 9x. Likewise on the B side, 8x of those elements should be thrown away. so now we have: \[1509x = 3n8x \iff 150x = 3n \iff 50\frac{x}{3} = n\] Now, to figure out what x is, we need to use the fact that the union of a group of sets contains every member of each set. if every element in S is repeated 10 times, that means every element in the union of the A's is repeated 10 times. This means that: \[\frac{150}{10} = 15\]is the number of elements in the the A's without repeats counted (same for the Bs as well). So now we have: \[50\frac{15}{3} = n \iff n = 45\]

sritama
 3 years ago
Best ResponseYou've already chosen the best response.0i understand.r u a teacher?
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